Question

When tert-butanol is placed in the presence of excess equimolar amounts of chloride and bromide ions in aqueous acid experimental data indicates 45% tert-butyl chloride and 55% tert-butyl bromide is obtained. On the other hand, when n-butanol is placed in the presence of excess equimolar amounts of chloride and bromide ions in aqueous acid experimental data indicates 16% of n-butyl chloride and 84% n-butyl bromide is obtained. Now when n-butanol is placed in the presence of excess equimolar amounts of chloride ions and bromide ions in acid dimethyl sulfoxide the experimental data provides an 84% n-butyl chloride and 16% n-butyl bromide as the outcome.

1.) write balanced chemical equations for all three reactions (hint: account for mole ratios!).

2.) draw complete mechanisms for all three reactions.

3.) calculate the theoretical yield for the chlorides and bromides if each reaction was started with 5.55ml of the corresponding alcohol, using an aqueous acid media.

Sample: tert

Theoretical g of chloride

Theoretical g of bromide

Sample: n-

Theoretical g of chloride

Theoretical g of bromide

Answer 1

1: The balanced chemical reactions are

Case-1:

100 (CH₃)₃COH + 45 Cl^- + 45 Br^- ----- > 45  (CH₃)₃CCl + 55 (CH₃)₃CBr

Case-2:

100 CH3CH2CH2CH2OH + 16 Cl^- + 84 Br^- ----- > 16 CH3CH2CH2CH2Cl + 84 CH3CH2CH2CH2Br

Case-3:

100 CH3CH2CH2CH2OH + 84 Cl^- + 16 Br^- ----- > 84 CH3CH2CH2CH2Cl + 16 CH3CH2CH2CH2Br

2: Mechanism:

Case-1: Since the participating alcohol is a tert-alcohol, this reaction proceeds via SN1 mechanism through carbocation formation.

Case-2:  Since the participating alcohol is a primary alcohol, this reaction proceeds via SN2 mechanism.

3.) We start with 5.55 mol of alcohol

Case-1:

100 (CH₃)₃COH + 45 Cl^- + 45 Br^- ----- > 45  (CH₃)₃CCl + 55 (CH₃)₃CBr

Theoritical mole of  (CH₃)₃CCl = (45 / 100) x 5.5 = 2.475 mol

Theoritical mole of  (CH₃)₃CBr = (55 / 100) x 5.5 = 3.025 mol

Case-2:

100 CH3CH2CH2CH2OH + 16 Cl^- + 84 Br^- ----- > 16 CH3CH2CH2CH2Cl + 84 CH3CH2CH2CH2Br

Theoritical mole of  CH3CH2CH2CH2Cl  = (16 / 100) x 5.5 = 0.88 mol

Theoritical mole of  CH3CH2CH2CH2Br  = (84 / 100) x 5.5 = 4.62 mol

Case - 3:

100 CH3CH2CH2CH2OH + 84 Cl^- + 16 Br^- ----- > 84 CH3CH2CH2CH2Cl + 16 CH3CH2CH2CH2Br

Theoritical mole of  CH3CH2CH2CH2Cl  = (84 / 100) x 5.5 = 4.62 mol

Theoritical mole of  CH3CH2CH2CH2Br  = (16 / 100) x 5.5 = 0.88 mol